## The Geometry of Solids

DG3CL592_10.qxd 7/1/02 11:05 AM Page 129 CONDENSED LESSON 10.1 The Geometry of Solids In this lesson, you ● ● Learn about polyhedrons, including ...
Author: Robert Marshall
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CONDENSED LESSON

10.1

The Geometry of Solids

In this lesson, you ● ●

Learn about polyhedrons, including prisms and pyramids Learn about solids with curved surfaces, including spheres, hemispheres, cylinders, and cones

In this chapter, you will study three-dimensional solid figures. Lesson 10.1 introduces various types of three-dimensional solids and the terminology associated with them. Read the lesson. Then, review what you read by completing the statements and answering the questions below. 1. A polyhedron is a solid formed by ________________ that enclose a single region of space. 2. A segment where two faces of a polyhedron intersect is called a(n) ________________. 3. A polyhedron with seven faces is called a ________________. 4. A tetrahedron has ________________ faces. 5. If each face of a polyhedron is enclosed by a regular polygon, and each face is congruent to the other faces, and the faces meet each vertex in exactly the same way, then the polyhedron is called a ________________. 6. A ________________ is a polyhedron with two faces, called bases, that are congruent, parallel polygons. 7. The faces of a prism that are not bases are called ________________. 8. What is the difference between a right prism and an oblique prism? 9. What type of solid is shown below? Be as specific as possible.

10. How many bases does a pyramid have? 11. The common vertex of a pyramid’s lateral faces is the _______________ of the pyramid. 12. What is the difference between an altitude of a pyramid and the height of a pyramid? 13. What type of solid is shown below? Be as specific as possible.

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Lesson 10.1 • The Geometry of Solids (continued) 14. Name three types of solids with curved surfaces. 15. A ________________ is the set of all points in space at a given distance from a given point. 16. A circle that encloses the base of a hemisphere is called a ________________. 17. Give an example of a real object that is shaped like a cylinder. Explain how you know it is a cylinder. 18. Tell which cylinder below is an oblique cylinder and which is a right cylinder. For each cylinder, draw and label the axis and the altitude.

19. The base of a cone is a ________________. 20. If the line segment connecting the vertex of a cone with the center of the base is perpendicular to the base, then the cone is a ________________.

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10.2

Volume of Prisms and Cylinders

In this lesson, you ● ● ●

Discover the volume formula for right rectangular prisms Extend the volume formula to cylinders and other right prisms Extend the volume formula to oblique prisms and cylinders

Volume is the measure of the amount of space contained in a solid. You use cubic units to measure volume: cubic inches in.3, cubic feet ft3, cubic yards yd3, cubic centimeters cm3, cubic meters m3, and so on. The volume of an object is the number of unit cubes that completely fill the space within the object.

1 1

1

1

Investigation: The Volume Formula for Prisms and Cylinders Find the volume of each right rectangular prism shown in Step 1 of your book. Try to come up with shortcuts so that you don’t have to count every cube. One strategy for finding the volume is to find the number of cubes in a “base” layer and then multiply by the number of layers. For example, if you consider the bottom of the prism in part b to be a base, then the base layer has 3(12), or 36 cubes. Because there are 8 layers, the total number of cubes is 36(8), or 288 cubes. Because each cube has a volume of 1 cm3, the volume is 288 cm3. Notice that the number of cubes in the base layer is the number of square units in the area of the base and that the number of layers is the height of the prism. So, you can find the volume of a right rectangular prism by multiplying the area of the base by the height. Conjecture A If B is the area of the base of a right rectangular prism and H is the height of the solid, then the formula for the volume is V  ________________.

C-88a

You can find the volume of any other right prism or cylinder the same way—simply multiply the area of the base by the height. For example, to find the volume of this cylinder, find the area of the circular base (the number of cubes in the base layer) and multiply by the height.

(continued)

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Lesson 10.2 • Volume of Prisms and Cylinders (continued) Extend Conjecture A to all right prisms and right cylinders. C-88b

Conjecture B If B is the area of the base of a right prism (or cylinder) and H is the height of the solid, then the formula for the volume is V  ________________. What about the volume of an oblique prism or cylinder? Page 516 of your book shows that you can model an oblique rectangular prism with a slanted stack of paper. You can then “straighten” the stack to form a right rectangular prism with the same volume. The right prism has the same base area and the same height as the original prism. So, you find the volume of the oblique prism by multiplying the area of its base by its height. A similar argument works for other oblique prisms and cylinders. Now, you can extend Conjecture B to oblique prisms and cylinders.

C-88c

Conjecture C The volume of an oblique prism (or cylinder) is the same as the volume of a right prism (or cylinder) that has the same ____________ and the same ________________. Be careful when calculating the volume of an oblique prism or cylinder. Because the lateral edges are not perpendicular to the bases, the height of the prism or cylinder is not the length of a lateral edge. You can combine Conjectures A–C into one conjecture.

C-88

Prism-Cylinder Volume Conjecture The volume of a prism or a cylinder is the ________________ multiplied by the ________________.

Examples A and B in your book show you how to find the volumes of a trapezoidal prism and an oblique cylinder. Read both examples. Try to find the volumes yourself before reading the solutions. The example below is Exercise 6 in your book.

EXAMPLE



Solution

6

12

The solid at right is a right cylinder with a 90° slice removed. Find the volume of the solid. Round your answer to two decimal places.

90 °

The base is 34 of a circle with radius 6 cm. The whole circle has area 36 cm2, so 34 of it has area 34(36), or 27 cm2. Now, use the volume formula. V  BH

The volume formula.

 27(12)

Substitute the area of the base and the height.

 324

Multiply.

 1017.88

Use the  key on your calculator to get an approximate answer.

The volume is about 1017.88 cm3.

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10.3

Volume of Pyramids and Cones

In this lesson, you ●

Discover the volume formula for pyramids and cones

There is a simple relationship between the volumes of prisms and pyramids with congruent bases and the same height, and between cylinders and cones with congruent bases and the same height.

Investigation: The Volume Formula for Pyramids and Cones If you have the materials listed at the beginning of the investigation, follow the steps in your book before reading on. The text below summarizes the results of the investigation. Suppose you start with a prism and a pyramid with congruent bases and the same height. If you fill the pyramid with sand or water and then pour the contents into the prism, it will fill the prism 13 of the way.

So, if you repeat the process, you will find that you can empty the contents of the pyramid into the prism three times, and the prism will be exactly full. You will get the same result no matter what shape the bases have, as long as the base of the pyramid is congruent to the base of the prism and the heights are the same. If you repeat the experiment with a cone and a cylinder with congruent bases and the same height, you will get the same result. That is, you can empty the contents of the cone into the cylinder exactly three times. The results can be summarized in a conjecture. Pyramid-Cone Volume Conjecture If B is the area of the base of a pyramid or a cone and H is the height of the solid, then the formula for the volume is V  13BH.

C-89

Example A in your book shows how to find the volume of a regular hexagonal pyramid. In the example, you need to use the 30°-60°-90° Triangle Conjecture to find the apothem of the base. Example B involves the volume of a cone. Work through both examples. Then, read the example on the next page. (continued)

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Lesson 10.3 • Volume of Pyramids and Cones (continued) EXAMPLE

Find the area of this triangular pyramid.

6 cm

45°



Solution

10 cm

The base is an isosceles right triangle. To find the area of the base, you need to know the lengths of the legs. Let l be the length of a leg, and use the Isosceles Right Triangle Conjecture. l 2  10 The length of the hypotenuse is the length of a leg times 2. 10 l    52 Solve for l. 2  The length of each leg is 52. Now, find the area of the triangle. 1 A  2bh Area formula for triangles. 1  25 2 5 2  Substitute the known values.  25

Multiply.

So, the base has area 25 cm2. Now, find the volume of the pyramid. 1 V  3BH Volume formula for pyramids and cones. 1  3(25)(6) Substitute the known values.  50

Multiply.

The volume of the pyramid is 50 cm3.

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10.4

Volume Problems

In this lesson, you ●

Use the volume formulas you have learned to solve problems

You have learned volume formulas for prisms, cylinders, pyramids, and cones. In this lesson, you will use these formulas to solve problems. In Example A in your book, the volume of a right triangular prism is given and you must find the height. In Example B, the volume of a sector of a right cylinder is given and you must find the radius of the base. Try to solve the problems yourself before reading the solutions. Below are some more examples.

EXAMPLE A



Solution

A swimming pool is in the shape of the prism shown at right. How many gallons of water can the pool hold? (A cubic foot of water is about 7.5 gallons.)

30 ft 14 ft 4 ft 16 ft

First, find the volume of the pool in cubic feet. The pool is in the shape of a trapezoidal prism. The trapezoid has bases of length 13 feet and 30 feet and a height of 14 feet. The height of the prism is 16 feet. V  BH 1  2(14)(13  30)  16  4816

13 ft

Volume formula for prisms. Use the formula 12h b1  b2 for the area of a trapezoid. Solve.

The pool has volume 4816 ft3. A cubic foot is about 7.5 gallons, so the pool holds 4816(7.5), or 36,120 gallons of water. The example below is Exercise 8 in your book.

EXAMPLE B

A sealed rectangular container 6 cm by 12 cm by 15 cm is sitting on its smallest face. It is filled with water up to 5 cm from the top. How many centimeters from the bottom will the water level reach if the container is placed on its largest face?

5 cm

15 cm

10 cm 6 cm



Solution

The smallest face is a 6-by-12-centimeter rectangle. When the prism is resting on its smallest face, the water is in the shape of a rectangular prism with base area 72 cm2 and height 10 cm. So, the volume of the water is 720 cm3.

12 cm

(continued)

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Lesson 10.4 • Volume Problems (continued) If the container is placed on its largest face, the volume will still be 720 cm3, but the base area and height will change. The area of the new base will be 12(15), or 180 cm2. You can use the volume formula to find the height. V  BH

Volume formula for prisms.

720  180H

Substitute the known values.

4H

Solve for H.

The height of the water will be 4 cm. So, the water level will be 4 cm from the bottom of the container.

EXAMPLE C



Solution

Find the volume of a rectangular prism with dimensions that are twice those of another rectangular prism with volume 120 cm3. For the rectangular prism with volume 120 cm3, let the dimensions of the rectangular base be x and y and the height be z. The volume of this prism is xyz, so xyz  120. The dimensions of the base of the other prism are 2x and 2y, and the height is 2z. Let V be the volume of this prism. Then, V  BH

Volume formula for prisms.

 (2x)(2y)(2z)

Substitute the known values.

 8xyz

Multiply.

 8(120)

Substitute the known value.

 960

Multiply.

The volume of the prism is 960 cm3.

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CONDENSED LESSON

10.5

Displacement and Density

In this lesson, you ● ●

Learn how the idea of displacement can be used to find the volume of an object Learn how to calculate the density of an object

To find the volumes of geometric solids, such as prisms and cones, you can use a volume formula. But what if you wanted to find the volume of an irregularly shaped object like a rock? As Example A in your book illustrates, you can submerge the object in a regularly shaped container filled with water. The volume of water that is displaced will be the same as the volume of the object. Read Example A carefully. Then, read the example below.

EXAMPLE A



Solution

When Tom puts a rock into a cylindrical container with diameter 7 cm, the water level rises 3 cm. What is the volume of the rock to the nearest tenth of a cubic centimeter? The “slice” of water that is displaced is a cylinder with diameter 7 cm and height 3 cm. Use the volume formula to find the volume of the displaced water. 3 cm

V  BH  (3.5)2

3

 115.5 The volume of the rock is about 115.5 cm3, the same as the volume of the displaced water.

7 cm

An important property of a material is its density. Density is the mass of matter in a given volume. It is calculated by dividing mass by volume. mass  density   volume The table on page 535 of your book gives the densities of ten metals. In Example B, the mass of a clump of metal and information about the amount of water it displaces are used to identify the type of metal. Read this example carefully. (continued)

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Lesson 10.5 • Displacement and Density (continued) The example below is Exercise 6 in your book.

EXAMPLE B



Solution

Chemist Dean Dalton is given a clump of metal and is told that it is sodium. He finds that the metal has a mass of 145.5 g. He places it into a nonreactive liquid in a square prism whose base measures 10 cm on each edge. If the metal is indeed sodium, how high should the liquid level rise? The table on page 535 of your book indicates that the density of sodium is 0.97 g/cm3. Use the density formula to find what the volume of the clump of metal should be if it is sodium. mass  Density formula. density   volume 145.5  Substitute the known information. 0.97   volume volume  0.97  145.5 145.5  volume   0.97 volume  150

Multiply both sides by volume. Divide both sides by 0.97. Simplify.

So, if the metal is sodium, it should displace 150 cm3 of water. Use the volume formula to find the height of the liquid that should be displaced. V  BH

Volume formula for prisms.

150  (10)(10)H

Substitute the known information. (The base is a 10-by-10 square.)

150  100H

Multiply.

1.5  H

Solve for H.

The liquid should rise 1.5 cm.

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10.6

Volume of a Sphere

In this lesson, you ●

Discover the volume formula for a sphere

You can find the volume of a sphere by comparing it to the volume of a cylinder. In the investigation you will see how.

Investigation: The Formula for the Volume of a Sphere If you have the materials listed at the beginning of the investigation, follow the steps in your book before reading on. The text below summarizes the results of the investigation. Suppose you have a hemisphere and a cylinder. The radius of the cylinder equals the radius of the hemisphere, and the height of the cylinder is twice the radius. Note that the cylinder is the smallest one that would enclose a sphere made from two of the hemispheres. The volume of the cylinder is

r 2(2r),

or

r

r

2r

2r 3.

If you fill the hemisphere with sand or water and empty the contents into the cylinder, the cylinder will be 13 full. If you fill the hemisphere again and empty the contents into the cylinder, the cylinder will be 23 full. So, the volume of the sphere (two hemispheres) is equal to 23 the volume of the cylinder. The volume of the cylinder is 2r 3. So, the volume of the sphere is 23(2r 3), or 43r 3. The results can be summarized as a conjecture. Sphere Volume Conjecture The volume of a sphere with radius r is given by the formula V  43r 3.

C-90

Read Example A in your book, which involves finding the percentage of plaster cut away when the largest possible sphere is carved from a cube. The solution involves four steps: Step 1

Find the volume of the sphere.

Step 2

Find the volume of the cube.

Subtract the volume of the sphere from the volume of the cube to find the volume of the plaster cut away. Step 3

To find the percentage cut away, divide the volume cut away by the volume of the original cube and convert the answer to a percent. Step 4

Try to find the solution to the problem in Example B on your own. Then, check your answer by reading the solution. (continued)

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Lesson 10.6 • Volume of a Sphere (continued) The following example is Exercise 6 in your book.

EXAMPLE

Find the volume of this solid.

18 cm 40°



Solution

The solid is a hemisphere with a 40° sector cut away. First, find the volume of the entire hemisphere. Because the formula for the volume of a sphere is V  43r 3, the formula for the volume of a hemisphere is V  23r 3. 2 V  3r 3 Volume formula for a hemisphere. 2  3(18)3 The radius is 18 cm.  3888

Simplify.

The volume of the entire hemisphere is 3888 cm3. A 40° sector has been cut 320 8   away, so the fraction of the hemisphere that remains is  360 , or 9 . So, the volume 8 of the solid is 9(3888), or 3456 cm3.

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10.7

Surface Area of a Sphere

In this lesson, you ●

Discover the formula for the surface area of a sphere

You can use the formula for the volume of a sphere, V  43r 3, to help you find the formula for the surface area of a sphere.

Investigation: The Formula for the Surface Area of a Sphere Imagine a sphere’s surface divided into tiny shapes that are nearly flat. The surface area of the sphere is equal to the sum of the areas of these “nearly polygons.” If you imagine radii connecting each of the vertices of the “nearly polygons” to the center of the sphere, you are mentally dividing the volume of the sphere into many “nearly pyramids.” Each of the “nearly polygons” is a base for one of the pyramids, and the radius, r, of the sphere is the height of the pyramid. The volume, V, of the sphere is the sum of the volumes of all the pyramids. B1

Step 1 Imagine that the surface of a sphere is divided into 1000 “nearly polygons” with areas B1, B2, B3, . . . , B1000. The surface area, S, of the sphere is the sum of the areas of these “nearly polygons”:

S  B1  B2  B3  · · ·  B1000 The pyramid with base area B1 has volume 13B1r, the pyramid with base area B2 has volume 13B2r, and so on. The volume of the sphere, V, is the sum of these volumes: 1 1 1 1 V  3 B1r  3 B2r  3 B3r  · · ·  3 B1000r Step 2

You can factor 13r from each term on the right side of the equation: 1 V  3r B1  B2  B3  · · ·  B1000 Because V  43r 3, you can substitute 43r 3 for V: 4 1 r 3  r B1  B2  B3  · · ·  B1000 3 3 Step 3

Now, substitute S for B1  B2  B3  · · ·  B1000: 4 1 r 3  rS 3 3 (continued)

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Lesson 10.7 • Surface Area of a Sphere (continued) Step 4

Solve the above equation for the surface area, S.

4 1 r 3  rS 3 3

The equation from Step 3.

4r 3  rS

Multiply both sides by 3.

4r 2  S

Divide both sides by r.

You now have a formula for finding the surface area of a sphere if you know the radius. You can state the result as a conjecture. C-91

Sphere Surface Area Conjecture The surface area, S, of a sphere with radius r is given by the formula S  4r 2.

The example in your book shows you how to find the surface area of a sphere if you know its volume. Try to find the surface area on your own before reading the solution. Then, try to solve the problem presented in the example below.

EXAMPLE

The base of this hemisphere has circumference 32 cm. Find the surface area of the hemisphere (including the base).

r



Solution

Because the base of the hemisphere has circumference 32 cm, the radius must be 16 cm. The area of the base of the hemisphere is (16)2, or 256 cm2. The area of the curved surface is half the surface area of a sphere with radius 16 cm. 1 S  24r 2 1  2(4)(16)2  512 So, the total surface area is 256  512, or 768 cm2.

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